For u-substitution of a definite integral, is it acceptable to have the limits (bounds) of the new integral equal to one another? For example, I have the problem:
$$
\int_{-1}^1x^4\left(1-x^2\right)^2dx
$$
I set $\,u=1-x^2\,$, which gives $\,x^2=1-u\,$, $\,x=\pm\sqrt{1-u}\,$ and $\,-\frac{1}{2}du=xdx$.
For the upper and lower bounds of the integral:
$$
x=-1 \rightarrow u=1-\left(-1\right)^2=1-1=0
$$
$$
x=1 \rightarrow u=1-\left(1\right)^2=1-1=0
$$
By using u-substitution, I arrive at
$$
-\frac{1}{2}\int_0^0\pm\sqrt{1-u}\left(1-u\right)u^2du
$$
We know that 
$$
\int_a^af(x)dx = 0
$$
I came to the conclusion that this integral is equal to zero as the upper and lower bounds are equal, however I know that that is not true.
I found other questions which address this subject, but none seem to make since to me as to why this is not true. 
I am aware that u-substitution is unnecessary for this solution, but I am unclear as to why it doesn't seem to work for $u=1-x^2$. I  suspect that the issue arises in the square root function, but I'm not sure, and would like to know how to avoid incorrect solutions like this in the future. 
Thank-you.
 A: 
As said by @JyrkiLahtonen in comment, you have to use the substitution which corresponds to the range of the variable of your integration.

The thing is that's why you're getting the value of integral 0, because $u$ has range from [0,1] and x has range from [-1,1]. 
This integration, if you don't want to make some substitution can be done by simple solving the question, by squaring and multiplying the contents.
You'll get 3 term with $x^n$ type and that will be easy to solve. I think the answer to this integral without limits will be
$$\frac {x^9} 9 - \frac {2x^7} 7 + \frac {x^5} 5$$
A: To use the substitution $u=1-x^2$ to evaluate,
$$
\int_{-1}^1x^4\left(1-x^2\right)^2\,\mathrm{d}x
$$
one has to remember that for $x\in[0,1]$, we have $x=\sqrt{1-u}$ giving $\mathrm{d}x=\frac{-\mathrm{d}u}{2\sqrt{1-u}}$ and for $x\in[-1,0]$, we have $x=-\sqrt{1-u}$ giving $\mathrm{d}x=\frac{\mathrm{d}u}{2\sqrt{1-u}}$. Thus, we get
$$
\int_{-1}^0x^4\left(1-x^2\right)^2\,\mathrm{d}x=\int_0^1(1-u)^2u^2\frac{\mathrm{d}u}{2\sqrt{1-u}}
$$
and
$$
\begin{align}
\int_0^1x^4\left(1-x^2\right)^2\,\mathrm{d}x
&=\int_{\color{#C00}{1}}^{\color{#C00}{0}}(1-u)^2u^2\frac{\color{#C00}{-}\mathrm{d}u}{2\sqrt{1-u}}\\
&=\int_0^1(1-u)^2u^2\frac{\mathrm{d}u}{2\sqrt{1-u}}
\end{align}
$$
A: $$I=\int_{-1}^1x^4(1-x^2)^2dx$$
notice that if we let $$f(x)=x^4(1-x^2)^2$$ then $$f(-x)=f(x)$$
and so the integral is equal to $$2\int_0^1x^4(1-x^2)^2dx$$
